### TUG OF WAR

How does the balance between gravitational force and forces opposing it shape the lives of stars? What happens when this balance is disturbed? We will try to answer the following questions in this article. We will follow the journey of a star from its youth to its old age and we won't stop there! We will also visit the graves and have a look at the dead bodies-the stellar remnants, and in our journey, we will witness spectacular powers of the universe at play-All of them the consequence of a tug of war!

### THE DAY THE SUN STOPPED SHINING

The historic solar eclipse of 1919 that Eddington used to finally prove Einstein's general theory of relativity, that established the end of the Newtonian era and went on to lay the foundation of cosmology and the understanding of the modern day Universe.

### THE APOLLO MISSION

On the 50th anniversary of the Apollo Mission, this article takes us back in history to the 1960s when humans were struggling to grasp with the idea of a world outside our own.

# Three Body Problem

Analysis of 3 Body problem is not very simple due to addition of various variables due to 3rd body.
3 body problem can’t be solved completely.

## Restricted 3 Body problem

Let one of the masses is infinitesimal small. ${m}_{3}\approx 0$$m_3\approx 0$
Now we have to solve equations only for ${m}_{3}$$m_3$. Because for ${m}_{1}$$m_1$ & ${m}_{2}$$m_2$, it is like 2 body problem
${m}_{3}\stackrel{¨}{\stackrel{\to }{{r}_{3}}}=-\frac{G{m}_{1}{m}_{3}\stackrel{\to }{{r}_{31}}}{{r}_{31}^{3}}-\frac{G{m}_{2}{m}_{3}\stackrel{\to }{{r}_{32}}}{{r}_{32}^{3}}$$m_3\ddot{\vec{r_3}}=-\frac{Gm_1m_3\vec{r_{31}}}{r_{31}^3}-\frac{Gm_2m_3\vec{r_{32}}}{r_{32}^3}$
$\stackrel{¨}{\stackrel{\to }{{r}_{3}}}=-\frac{G{m}_{1}\stackrel{\to }{{r}_{31}}}{{r}_{31}^{3}}-\frac{G{m}_{2}\stackrel{\to }{{r}_{32}}}{{r}_{32}^{3}}$$\ddot{\vec{r_3}}=-\frac{Gm_1\vec{r_{31}}}{r_{31}^3}-\frac{Gm_2\vec{r_{32}}}{r_{32}^3}$
Using Cartesian Coordinates
$\stackrel{¨}{x}=-\frac{G{m}_{1}\left(x-{x}_{1}\right)}{\sqrt{\left(x-{x}_{1}{\right)}^{2}+\left(y-{y}_{1}{\right)}^{2}+\left(z-{z}_{1}{\right)}^{2}}}$$\ddot{x}=-\frac{Gm_1(x-x_1)}{\sqrt{(x-x_1)^2+(y-y_1)^2+(z-z_1)^2}}$
$\stackrel{¨}{y}=-\frac{G{m}_{1}\left(y-{y}_{1}\right)}{\sqrt{\left(x-{x}_{1}{\right)}^{2}+\left(y-{y}_{1}{\right)}^{2}+\left(z-{z}_{1}{\right)}^{2}}}$$\ddot{y}=-\frac{Gm_1(y-y_1)}{\sqrt{(x-x_1)^2+(y-y_1)^2+(z-z_1)^2}}$
$\stackrel{¨}{z}=-\frac{G{m}_{1}\left(z-{z}_{1}\right)}{\sqrt{\left(x-{x}_{1}{\right)}^{2}+\left(y-{y}_{1}{\right)}^{2}+\left(z-{z}_{1}{\right)}^{2}}}$$\ddot{z}=-\frac{Gm_1(z-z_1)}{\sqrt{(x-x_1)^2+(y-y_1)^2+(z-z_1)^2}}$

## Circular Restricted 3 body problem

Suppose that the first two masses, ${m}_{1}$$m_1$ and ${m}_{2}$$m_2$, execute circular orbits about their common center of mass.

Let us define a Cartesian coordinate system $\left(\xi ,\phantom{\rule{thinmathspace}{0ex}}\eta ,\phantom{\rule{thinmathspace}{0ex}}\zeta \right)$$(\xi,\,\eta,\,\zeta)$ in an inertial reference frame whose origin coincides with the center of mass, $C$$C$, of the two orbiting masses. Let the orbital plane of these masses coincide with the $\xi -\eta$$\xi-\eta$ plane, and let them both lie on the $\xi$$\xi$-axis at time $t=0$$t=0$–see Figure. Suppose that $R$$R$ is the constant distance between the two orbiting masses, ${r}_{1}$$r_1$ the constant distance between mass ${m}_{1}$$m_1$ and the origin, and ${r}_{2}$$r_2$ the constant distance between mass ${m}_{2}$$m_2$ and the origin. Moreover, let $\omega$$\omega$ be the constant orbital angular velocity.

$\left(\begin{array}{c}x\\ y\\ z\end{array}\right)=\left(\begin{array}{ccccc}\mathrm{cos}\omega t& & -\mathrm{sin}\omega t& & 0\\ \mathrm{sin}\omega t& & \mathrm{cos}\omega t& & 0\\ 0& & 0& & 1\end{array}\right)\left(\begin{array}{c}\xi \\ \eta \\ \zeta \end{array}\right)$$% $
Differentiate it twice put it in cartesian equations of motion of $m_3$. We get
$\stackrel{¨}{\xi }-2\omega \eta =\frac{\mathrm{\partial }\mathbb{U}}{\mathrm{\partial }\xi }\cdots \cdots \left(7\right)$$\ddot{\xi}-2\omega\eta=\frac{\partial\Bbb{U}}{\partial\xi}\cdots\cdots(7)$
$\stackrel{¨}{\eta }+2\omega \xi =\frac{\mathrm{\partial }\mathbb{U}}{\mathrm{\partial }\eta }\cdots \cdots \left(8\right)$$\ddot{\eta}+2\omega\xi=\frac{\partial\Bbb{U}}{\partial\eta}\cdots\cdots(8)$
$\stackrel{¨}{\zeta }=\frac{\mathrm{\partial }\mathbb{U}}{\mathrm{\partial }\zeta }\cdots \cdots \left(9\right)$$\ddot{\zeta}=\frac{\partial\Bbb{U}}{\partial\zeta}\cdots\cdots(9)$

$\left(7\right)×\stackrel{˙}{\xi }+\left(8\right)×\stackrel{˙}{\eta }+\left(9\right)×\stackrel{˙}{\zeta }$

$\stackrel{˙}{\xi }\stackrel{¨}{\xi }+\stackrel{˙}{\eta }\stackrel{¨}{\eta }+\stackrel{˙}{\zeta }\stackrel{¨}{\zeta }=\stackrel{˙}{\xi }\frac{\mathrm{\partial }\mathbb{U}}{\mathrm{\partial }\xi }+\stackrel{˙}{\eta }\frac{\mathrm{\partial }\mathbb{U}}{\mathrm{\partial }\eta }+\stackrel{˙}{\zeta }\frac{\mathrm{\partial }\mathbb{U}}{\mathrm{\partial }\zeta }\cdots \cdots \left(10\right)$$\dot{\xi}\ddot{\xi}+\dot{\eta}\ddot{\eta}+\dot{\zeta}\ddot{\zeta}=\dot{\xi}\frac{\partial\Bbb{U}}{\partial\xi}+\dot{\eta}\frac{\partial\Bbb{U}}{\partial\eta}+\dot{\zeta}\frac{\partial\Bbb{U}}{\partial\zeta}\cdots\cdots(10)$
if $\mathbb{U}=\mathbb{U}\left(\xi ,\phantom{\rule{thinmathspace}{0ex}}\eta ,\phantom{\rule{thinmathspace}{0ex}}\zeta \right)$$\Bbb{U}=\Bbb{U}(\xi,\,\eta,\,\zeta)$
$⇒\frac{d\mathbb{U}}{dt}=\frac{\mathrm{\partial }\mathbb{U}}{\mathrm{\partial }\xi }\cdot \frac{\mathrm{\partial }\xi }{\mathrm{\partial }t}+\frac{\mathrm{\partial }\mathbb{U}}{\mathrm{\partial }\eta }\cdot \frac{\mathrm{\partial }\eta }{\mathrm{\partial }t}+\frac{\mathrm{\partial }\mathbb{U}}{\mathrm{\partial }\zeta }\cdot \frac{\mathrm{\partial }\zeta }{\mathrm{\partial }t}$$\Rightarrow \frac{d\Bbb{U}}{dt}=\frac{\partial\Bbb{U}}{\partial\xi}\cdot\frac{\partial\xi}{\partial t}+\frac{\partial\Bbb{U}}{\partial\eta}\cdot\frac{\partial\eta}{\partial t}+\frac{\partial\Bbb{U}}{\partial\zeta}\cdot\frac{\partial\zeta}{\partial t}$
It is R.H.S. of $\left(10\right)$$(10)$. Also
$\frac{d}{dt}{\stackrel{˙}{\xi }}^{2}=2\stackrel{˙}{\xi }\stackrel{¨}{\xi }$$\frac{d}{dt}{\dot{\xi}}^2=2\dot{\xi}\ddot{\xi}$
eqn $(10)$ becomes
$\frac{1}{2}\frac{d}{dt}\left({\stackrel{˙}{\xi }}^{2}+{\stackrel{˙}{\eta }}^{2}+{\stackrel{˙}{\zeta }}^{2}\right)=\frac{d\mathbb{U}}{dt}$$\frac{1}{2}\frac{d}{dt}({\dot{\xi}}^2+{\dot{\eta}}^2+{\dot{\zeta}}^2)=\frac{d\Bbb{U}}{dt}$
We know ${\dot{\xi}}^2+{\dot{\eta}}^2+{\dot{\zeta}}^2=v^2$.
By Jacobi Integral (External Link) - For Extra Information
${v}^{2}=2\mathbb{U}-C$$v^2=2\Bbb{U}-C$
For ${v}^{2}\ge 0,2\mathbb{U}\ge C$$v^2\ge 0, 2\Bbb{U}\ge C$. This gives
$\mathbb{U}=\frac{{\omega }^{2}}{2}\left({\xi }^{2}+{\eta }^{2}\right)+\frac{{\mu }_{1}}{{r}_{1}}+\frac{{\mu }_{2}}{{r}_{2}}$$\Bbb{U}=\frac{\omega^2}{2}(\xi^2+\eta^2)+\frac{\mu_1}{r_1}+\frac{\mu_2}{r_2}$
If I term is large
${\xi }^{2}+{\eta }^{2}\ge \frac{2C}{{\omega }^{2}}$$\xi^2+\eta^2\ge \frac{2C}{\omega^2}$
Path would be enclosing ${m}_{1}$$m_1$ and ${m}_{2}$$m_2$. On changing $C$$C$ we get different Lagrangian Points